|
|
见Kerstin Jordaan, Properties of orthogonal polynomial, Lecture-1 Lecture-2 Lecture-3
Show that Roots of Orthogonal Polynomials are interlaced.
interlacing properties of zeros of orthogonal polynomials
Orthogonal Polynomials
Zeros of classical orthogonal polynomials
If $\left\{p_n(x)\right\}_{n=0}^{\infty}$ is a sequence of orthogonal polynomials on the interval $(a, b)$ with respect to the weight function $w(x)$, then the polynomial $p_n(x)$ has exactly $n$ real simple zeros in the interval $(a, b)$.
Since $\operatorname{deg}\left(p_n\right)=n$ the polynomial has at most $n$ real zeros.
Suppose that $p_n(x)$ has $m \leq n$ distinct real zeros $x_1, x_2, \ldots, x_m$ in $(a, b)$ of odd order (or multiplicity).
Then the polynomial
\[
p_n(x)\left(x-x_1\right)\left(x-x_2\right) \ldots\left(x-x_m\right)
\]
does not change sign on the interval $(a, b)$.
This implies that
\[
\int_a^b w(x) p_n(x)\left(x-x_1\right)\left(x-x_2\right) \ldots\left(x-x_m\right) d x \neq 0 .
\]
By orthogonality this integral equals zero if $m<n$.
Hence: $m=n$, which implies that $p_n(x)$ has $n$ distinct real zeros of odd order in $(a, b)$.
This proves that all $n$ zeros are distinct and simple (have order or multiplicity equal to one).
An important consequence of the recurrence relation, is the Christoffel-Darboux formula (See Assignment 1, Ex. 3).
Theorem
A sequence of orthogonal polynomials $\left\{p_n(x)\right\}_{n=0}^{\infty}$ satisfies
\[
\sum_{k=0}^n \frac{p_k(x) p_k(y)}{h_k}=\frac{k_n}{h_n k_{n+1}} \frac{p_{n+1}(x) p_n(y)-p_{n+1}(y) p_n(x)}{x-y}, n=0,1,2, \ldots\tag2
\]
and
\[
\sum_{k=0}^n \frac{\left\{p_k(x)\right\}^2}{h_k}=\frac{k_n}{h_n k_{n+1}}\left(p_{n+1}^{\prime}(x) p_n(x)-p_{n+1}(x) p_n^{\prime}(x)\right), n=0,1,2, \ldots\tag3
\]
(2) is called the Christoffel-Darboux formula and (3) its confluent form.
(3) yields another important property of zeros of orthogonal polynomials.
Interlacing of zeros
Theorem
If $\left\{p_n(x)\right\}_{n=0}^{\infty}$ is a sequence of orthogonal polynomial on the interval $(a, b)$ with respect to the weight function $w(x)$, then the zeros of $p_n(x)$ and $p_{n+1}(x)$ separate each other.
$f\left\{x_{n, k}\right\}_{k=1}^n$ and $\left\{x_{n+1, k}\right\}_{k=1}^{n+1}$ denote the consecutive zeros of $p_n(x)$ and $p_{n+1}(x)$ respectively, then we have
\[
a<x_{n+1,1}<x_{n, 1}<x_{n+1,2}<x_{n, 2}<\ldots<x_{n+1, n}<x_{n, n}<x_{n+1, n+1}<b .
\]
Proof
Note that
\[
h_n=\int_a^b w(x)\left\{p_n(x)\right\}^2 d x>0, n=0,1,2, \ldots
\]
This implies that
\[
\frac{k_n}{h_n k_{n+1}}\left(p_{n+1}^{\prime}(x) p_n(x)-p_{n+1}(x) p_n^{\prime}(x)\right)=\sum_{k=0}^n \frac{\left\{p_k(x)\right\}^2}{h_k}>0
\]
Hence
\[
\frac{k_n}{k_{n+1}}\left(p_{n+1}^{\prime}(x) p_n(x)-p_{n+1}(x) p_n^{\prime}(x)\right)>0
\]
Now suppose that $x_{n+1, k}$ and $x_{n+1, k+1}, k=1,2, \ldots, n$ are any two consecutive zeros of $p_{n+1}(x)$ with $x_{n+1, k}<x_{n+1, k+1}$.
Since all $n+1$ zeros of $p_{n+1}(x)$ are real and simple, it follows from Rolle's theorem that $p_{n+1}^{\prime}\left(x_{n+1, k}\right)$ and $p_{n+1}^{\prime}\left(x_{n+1, k+1}\right)$ have opposite signs.
Hence we have
\[
p_{n+1}\left(x_{n+1, k}\right)=0=p_{n+1}\left(x_{n+1, k+1}\right)\quad \text { and }\quad p_{n+1}^{\prime}\left(x_{n+1, k}\right) p_{n+1}^{\prime}\left(x_{n+1, k+1}\right)<0
\]
This implies that $p_n\left(x_{n+1, k}\right) p_n\left(x_{n+1, k+1}\right)<0$. Why?
It follows from the continuity of $p_n(x)$ that there should be at least one zero of $p_n(x)$ between $x_{n+1, k}$ and $x_{n+1, k+1}$
This holds for each pair of consecutive zeros of $p_{n+1}(x)$ so there is exactly one zero of $p_n(x)$ between $x_{n+1, k}$ and $x_{n+1, k+1}$.
|
|
